Thread: Chance of pairs, rolling 5 dices simultaneously

Hello all,

Can you please help me solve the following problem?
This is from A first course in probability by Seldon Ross 7th edition.

Chapter 2 Problem 16
Poker dice is played by simultaneously rolling 5 dice. Show that:
a) P { no two alike } = 0.0926
b) P { one pair } = 0.4630
c) P { two pair } = 0.2315
d) P { three alike } = 0.1543
e) P { full house } = 0.0386
f) P { four alike } = 0.0193
g) P { five alike } = 0.0008

Thanks!

Ok I can solve problems a,b and g.

a) P { no two alike } = = 0.0926


b) P { one pair } = = 0.4630


g) P { five alike } = = 0.0008

However the trick I used for b doesn't work for c,d,e and f, and I don't understand
why? This is how I solve b:

The numerical value for the pair can be chosen in 6 ways and can be placed in ways, and then there are 3 'slots' left. The numerical value for the remaining 3 slots can be chosen in 5*4*3 ways. The whole thing is divided by the total number of possibilities which is .

But as I said this same line of reasoning doesn't work for c,d,e and f? Am I doing something wrong?

Originally Posted by desperate87

Poker dice is played by simultaneously rolling 5 dice. Show that:
c) P { two pair } = 0.2315
d) P { three alike } = 0.1543
e) P { full house } = 0.0386
f) P { four alike } = 0.0193

It is just counting. All the dominators .

c) Pick the two pairs: . Pick the odd die out . Now scramble.


You try the rest!

Thank you Plato
I finally understand it.

I'll post my results when I'm done